The number of nuts in a can of mixed nuts is found to be normally distributed, with a mean of 500 nuts and a standard deviation
of 20 nuts. My can of mixed nuts has only 455 nuts. What is the z-score for this can of nuts?
2 answers:
Answer:
z = -2.25
Step-by-step explanation:
Answer:
-2.25
Step-by-step explanation:
We have that the mean (m) is equal to 500, the standard deviation (sd) 20
They ask us for P (x <455)
For this, the first thing is to calculate z, which is given by the following equation:
z = (x - m) / sd
We have all these values, replacing we have:
z = (455 - 500) / (20)
z = -2.25
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Step-by-step explanation:
Answer:
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Step-by-step explanation:
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